Second order obstacle problem

Imagine a circular (or cylindrical, in cross-section) object being supported by an elastic string. Like this:

Obstacle problem

To actually compute the equilibrium mass-string configuration, I would have to take some values for the mass of the object and for the resistance of the string. Instead, I simply chose the position of the object: it is the unit circle with center at . It remains to find the equilibrium shape of the string. The shape is described by equation where minimizes the appropriate energy functional subject to boundary conditions and the obstacle. The functional could be the length

or its quadratization

The second one is nicer because it yields linear Euler-Lagrange equation/inequality. Indeed, the obstacle permits one-sided variations with smooth and compactly supported. The linear term of is , which after integration by parts becomes . Since the minimizer satisfies , the conclusion is whenever . Therefore, everywhere (at least in the sense of distributions), which means is a convex function. In the parts where the string is free, we can do variation of either sign and obtain ; that is, is an affine function there.

The convexity of in the part where it touches the obstacle is consistent with the shape of the obstacle: the string can assume the same shape as the obstacle.

The function can now be determined geometrically: the only way the function can come off the circle, stay convex, and meet the boundary condition is by leaving the circle along the tangents that pass through the endpoint . This is the function pictured above. Its derivative is continuous: Lipschitz continuous, to be precise.

First derivative is Lipschitz continuous

The second derivative does not exist at the transition points. Still, the minimizer has a higher degree of regularity (Lipschitz continuous derivative) than a generic element of the function space in which minimization takes place (square-integrable derivative).

As a bonus, the minimizer of energy turns out to minimize the length as well.

All in all, this was an easy problem. Next post will be on its fourth-order version.